數學功課.要有步驟ar...help ME..急!!
要有步驟ar..
1.(6a+5)+(2a-4)
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2.(y^2+3)+(2y^2-7)
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3.(2d-3d^2)-(d^2+3d)
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回答 (2)
✔ 最佳答案
1.(6a+5)+(2a-4)
=6a+2a+5-4
=8a+1
2.(y²+3)+(2y²-7)
=y²+2y²+3-7
=3y²-4
3.(2d-3d²)-(d²+3d)
=2d-3d²-d²-3d
=2d-3d-3d²-d²
=-d-4d²
1.(6a+5)+(2a-4)
=6a+5+2a-4
=8a+1
2..(y^2+3)+(2y^2-7)
=y^2+3+2y^2-7
=3y^2-4
3.(2d-3d^2)-(d^2+3d)
=.2d-3d^2-d^2-3d
=-3d^2-d^2-3d+2d
=-4d^2-d
參考: me
收錄日期: 2021-04-13 14:21:08
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